4.0 moles of `PCI_5` dissociated at 760 K in a 2 litre flask `PCI_5 (g) hArr PCI_3 (g) +CI_2(g)` at equilibrium .
0.8 mole of `CI_2` was present in the flask .The equilibrium constant would be

Options

A. `1.0 xx 10^(-1)`
B. `1.0 xx 10^(-4)`
C. `1.0xx10^(-2)`
D. `1.0 xx 10^(-3)`

Answer

To find the equilibrium constant (K) for the dissociation of PCl5, we can follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of phosphorus pentachloride (PCl5) can be represented as:
\[ \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (g) + \text{Cl}_2 (g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- Moles of PCl5 = 4.0 moles
- Moles of PCl3 = 0 moles
- Moles of Cl2 = 0 moles

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the amount of PCl5 that dissociates. At equilibrium, the moles will be:
- Moles of PCl5 = \( 4 - x \)
- Moles of PCl3 = \( x \)
- Moles of Cl2 = \( x \)

### Step 4: Use the information given
We know that at equilibrium, there are 0.8 moles of Cl2 present. Therefore, we can set:
\[ x = 0.8 \, \text{moles of Cl2} \]

### Step 5: Calculate the moles of PCl5 and PCl3 at equilibrium
Using \( x = 0.8 \):
- Moles of PCl5 = \( 4 - 0.8 = 3.2 \)
- Moles of PCl3 = \( 0.8 \)

### Step 6: Calculate the concentrations
Since the volume of the flask is 2 liters, we can calculate the concentrations:
- Concentration of PCl5:
\[ [\text{PCl}_5] = \frac{3.2 \, \text{moles}}{2 \, \text{liters}} = 1.6 \, \text{M} \]
- Concentration of PCl3:
\[ [\text{PCl}_3] = \frac{0.8 \, \text{moles}}{2 \, \text{liters}} = 0.4 \, \text{M} \]
- Concentration of Cl2:
\[ [\text{Cl}_2] = \frac{0.8 \, \text{moles}}{2 \, \text{liters}} = 0.4 \, \text{M} \]

### Step 7: Write the expression for the equilibrium constant
The equilibrium constant \( K \) for the reaction is given by:
\[ K = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]

### Step 8: Substitute the concentrations into the equilibrium expression
Substituting the values we calculated:
\[ K = \frac{(0.4)(0.4)}{1.6} = \frac{0.16}{1.6} = 0.1 \]

### Step 9: Final answer
Thus, the equilibrium constant \( K \) is:
\[ K = 1.0 \times 10^{-1} \]